Question: Let $f(x)=\sqrt{4x-3}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $1\leq x\leq3$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1.5$ (Choice B) B $1.75$ (Choice C) C $2$ (Choice D) D $2.25$
Solution: According to the Mean Value Theorem, there exists a number $c$ in the open interval $1<x<3$ such that $f'(c)$ is equal to the average rate of change of $f$ over the interval: $f'(c)=\dfrac{f(3)-f(1)}{(3)-(1)}$ First, let's find that average rate of change: $\dfrac{f(3)-f(1)}{(3)-(1)}=\dfrac{3-1}{2}={1}$ Now, let's differentiate $f$ and find the $x$ -value for which $f'(x)={1}$. $f'(x)=\dfrac{2}{\sqrt{4x-3}}$ The solution of $f'(x)=1$ is $x=1.75$. $x=1.75$ is indeed within the interval $1<x<3$. In conclusion, $c=1.75$.